Higher June 2025 Paper 6 Q22
22 The diagram shows a square-based pyramid ABCDE.
O is the centre of the base.
The length of the base is 10.7 cm.
Angle EAO is 67°.
Calculate the length of AE.
You must show your working. [6]
| Answer | Marks | Part marks and guidance | |||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 19.3[4…] to 19.3[7…] or 19.4 with correct working | 6 | M2 for \(\left[\frac{1}{2} \times\right]\sqrt{10.7^2 + 10.7^2}\) or for \(\sqrt{5.35^2 + 5.35^2}\) or for \(\left[\frac{1}{2} \times\right] \frac{10.7}{\cos 45}\) or \(\left[\frac{1}{2} \times\right] \frac{10.7}{\sin 45}\) or for \(\frac{5.35}{\cos 45}\) or \(\frac{5.35}{\sin 45}\) or M1 for \(10.7^2 + 10.7^2\) or \(5.35^2 + 5.35^2\) or for \(\cos 45 = \frac{10.7}{\text{AC}}\) oe or \(\sin 45 = \frac{10.7}{\text{AC}}\) oe or for \(\cos 45 = \frac{5.35}{\text{AO}}\) oe or \(\sin 45 = \frac{5.35}{\text{AO}}\) oe | Correct working requires evidence of at least M1 AND M1 Accept intermediate values rot to at least 3sf. If M1A1 seen then award M2A1
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| A1 for 7.56 to 7.57 or 7.6 AND | |||||||||||||||||
| M2 for \(\frac{\textit{their } 7.566\ldots}{\cos 67}\) or \(\frac{\textit{their } 7.566\ldots}{\sin 23}\) or \(\frac{\text{AE}}{\sin 90} = \frac{\textit{their } 7.566\ldots}{\sin 23}\) or M1 for \(\cos 67 = \frac{\textit{their } 7.566\ldots}{\text{AE}}\) oe or \(\sin 23 = \frac{\textit{their } 7.566\ldots}{\text{AE}}\) oe or \(\frac{\sin 90}{\text{AE}} = \frac{\sin 23}{\textit{their } 7.566}\) oe | their 7.566…must follow from the award of the first M2 | ||||||||||||||||
| If 0, 1 or 2 scored, instead award SC3 for answer 19.3 to 19.4 with no or insufficient working If 0 or M1 scored, instead award SC2 for 7.56 to 7.57 or 7.6 with no or insufficient working If 0 scored instead award SC1 for 15.1[3…] or 57.2… with no or insufficient working | Mark less efficient methods by considering how many steps remain to reach the correct answer and match against the most comparable stage: eg Assuming AO has been found, then using trig to find EO is unnecessary and inefficient, so no further marks are awarded for just finding EO and stopping. However, if EO is then used in a correct Pythagoras expression with AO or a correct trig expression with 67° that lead directly or indirectly to finding AE, then M2 or M1 may be awarded. | ||||||||||||||||